Question

Question #eca11

Answer 1

`int_0^oo2/(1+x^2)dx=pi`

Explanation:

We will use the following:

• `int1/(1+x^2)dx = arctan(x)+C`
• `arctan(0) = 0`
• `lim_(x->oo)arctan(x) = pi/2`

With that:

`int_0^oo2/(1+x^2)dx = lim_(N->oo)2int_0^N1/(1+x^2)dx`

`=lim_(N->oo)2[arctan(x)]_0^N`

`=lim_(N->oo)2(arctan(N)-arctan(0))`

`=lim_(N->oo)2arctan(N)`

`=2(pi/2)`

`=pi`